Fix: Minor indentation Changes

Recursion and Backtracking/1.md

@@ -23,7 +23,7 @@ __Intuition__
 
 The  main idea is to represent a problem in terms of one or more smaller problems, and add one or more base conditions that stop the recursion. 
 _For example_, we compute factorial n if we know factorial of (n-1). The base case for factorial would be n = 0. We return 1 when n = 0.
--- --
+
 
 Power
 -----
@@ -35,7 +35,7 @@ def pow(n, k):
     if k == 0: return 1
     return n*pow(n, k - 1)
 ```
-__Time Complexity__: $O(n)$
+__Time Complexity__: O(n)
 
 __Optimised solution:__
 ```python
@@ -54,7 +54,7 @@ To allow reuse of answers.
 
 <img src="https://user-images.githubusercontent.com/35702912/66316190-d30e1f00-e934-11e9-8089-85c6dc69baa7.jpg" data-canonical-src="https://user-images.githubusercontent.com/35702912/66316190-d30e1f00-e934-11e9-8089-85c6dc69baa7.jpg" width="500" />
 
-__Time Complexity__ (assuming all multiplications are O(1))? $O(\log_2 k)$
+__Time Complexity__ (assuming all multiplications are O(1))? O(\log_2 k)$O(\log_{2}k)$
 
 
 Break it into 3 parts? k//3 and take care of mod1 and mod2.
@@ -71,7 +71,7 @@ The idea is to consider two cases for every element.
 (i) Consider current element as part of current subset.
 (ii) Do not consider current element as part of current subset.
 
-Number of subsets? $2^n$
+Number of subsets? $2^{n}$
 
 Explain that we want combinations, and not permutations. [1, 4] = [4, 1]
 
@@ -164,7 +164,7 @@ def subsets(A, i, aux, p):
     no_take = subsets(A, i+1, aux, False)
 ```
 
-__Time Complexity__: $O(2^n)$
+__Time Complexity__: $O(2^n)$ <br>
 __Space Complexity__: $O(n^2)$, because we're creating new aux arrays.
 
 -- --
@@ -228,7 +228,7 @@ def subsetSum2(A,N,cur_sum, i, target):
     no_take = subsetSum2(A,N,cur_sum, i+1, target)
     return take + no_take
 ```
-__Time Complexity__ :  $O(2 ** (Target/MinElement))$
+__Time Complexity__ :  $O(2 ** (Target/MinElement))$ <br>
 __Space Complexity__:  $O(Target/Min Element)$